When viewing a piece of art that is behind glass, one often is affected by the light that is reflected off the front of the glass (called gl

Question

When viewing a piece of art that is behind glass, one often is affected by the light that is reflected off the front of the glass (called glare), which can make it difficult to see the art clearly. One solution is to coat the outer surface of the glass with a thin film to cancel part of the glare. Part A If the glass has a refractive index of 1.58 and you use TiO2, which has an index of refraction of 2.62, as the coating, what is the minimum film thickness that will cancel light of wavelength 480 nm

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Vân Khánh 5 months 2021-08-16T04:37:58+00:00 1 Answers 6 views 0

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    2021-08-16T04:39:27+00:00

    Answer:

      t = 9.16 10⁻⁸ m

    Explanation:

    In this exercise we fear a case of interference by reflection,

    We have two aspects

    * when the light ray passes from the air to the TiO2 that has a higher refraction start, the reflected ray has a phase change of 180º

    * when the beam is inside the material its wavelength changes

                λₙ = λ₀ / n

    with these two done the equation for destructive interference remains

               2t = m  λₙ

               2nt = m λ₀

    They ask us for the thickness t of the film for m = 1

               t = m λ₀ / 2n

    let’s calculate

              t = \frac{1 \ 480 1 10^9 }{2 \ 2.62}

              t = 9.16 10⁻⁸ m

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